A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #7/15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2017-08-11T19:45:10

0.0115 m (assuming #g = 10 m/s^2#

The maximum height the projectile reaches is

#h = {u^2 sin^2 theta}/{2 g} =0.01016 m#

The horizontal distance it travels up to this point is

#d = {u^2 sin theta cos theta}/g = 0.00544 m#

The distance of this point from the starting point is

#sqrt{h^2+d^2} = 0.0115 m$